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### Network # Confusion over inductors and cemf

J

#### James W

Jan 1, 1970
0
Consider a simple inductive cicuit with a 1v(p-to-p) AC source at
1Hz,and an inductor with Z=1ohm.

The inductors value would be 1ohm=2*pi*1Hz*L, so L= 1/2pi

Looking at the standard Voltage and Current drawings, we see current
lagging voltage by 90degrees.

Here's my problem. At 90 degrees, the applied voltage is 1volt. The
current is zero. di/dt is 1, and we know that the cemf V=(di/dt)L=1/2p
Volts.

So... why if applied V is 1v and cemf is less than 1v do we have a
current of zero?

I've got the sneaky feeling that I'm trying to add apples and oranges.

My assumption was that the reason the current lags the voltage is the
back emf is being added to ( subtracted from ) the applied voltage to
give us the instantaneous voltage that then drives the instantenous
current, but perhaps this whole theory is wrong.

HELP!!

R

#### Robert Monsen

Jan 1, 1970
0
James W said:
Consider a simple inductive cicuit with a 1v(p-to-p) AC source at
1Hz,and an inductor with Z=1ohm.

The inductors value would be 1ohm=2*pi*1Hz*L, so L= 1/2pi

Looking at the standard Voltage and Current drawings, we see current
lagging voltage by 90degrees.

Here's my problem. At 90 degrees, the applied voltage is 1volt. The
current is zero. di/dt is 1, and we know that
^^^^^^^^^^
Why do you think di/dt is 1? Its 2pi.

J

#### James W

Jan 1, 1970
0
Yep.. that's where I was going wrong. Thanks
- jim

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